Question

A lake contains V =2 x 105 m3 of water and is fed by a stream discharging Qin = 9 x 104 m3/year with a contaminant concentration of C = 6 mg/L. Evaporation across the surface takes away about 1 x 104 m3/year of water. If we assume that evaporation is the only means of water loss in the lake and it is at steady state what is the downstream flow from the lake? If the pollutant decays in the lake at k = 0.10/year what is the downstream concentration of the contaminant?

Answer 1

Answer:

Qdown = 0 m³/year

C = 27 mg/L

Step-by-step explanation:

Given

V = 2*10⁵ m³

Qin = 9*10⁴ m³/year

Cin = 6 mg/L

Qevap = 10⁴ m³/year

K = 0.10/year

Qdown = ?

C = ?

Take volume V of the entire lake as  the control volume.  If we assume that evaporation is the only means of water loss in the lake and it is at steady state (= situation unchanging over time) we have

Qdown = 0 m³/year

Budget reduces to:

0 = Qin*Cin - K*V*C

⇒  C = (Qin*Cin)/(K*V)

⇒  C = (9*10⁴ m³/year*6 mg/L)/(0.10/year*2*10⁵ m³) = 27 mg/L

Answer 2

Answer:

$Q_{down}=8 * 10^4 \:\:m^3/year$

$C_{down}=5.4\:\:mg/L$

Step-by-step explanation:

$Q_{down}=Q_{in}-Q_{evap}=9 * 10^4-1 * 10^4=8 * 10^4 \:\:m^3/year$

$V\frac{dC}{dt} =Q_{up}C_{up}-Q_{evap}C_{evap}-Q_{down}C_{down}-kVC\\\\where \:\:\:C_{down}=?,\:C_{up}=C=6\:\:mg/L\:\:\:and\:\:\:\frac{dC}{dt}=0,\:C_{evap}=0$

$0=Q_{up}C-(Q_{down}+kV)C_{down}\\where\:\:\:V=2*10^5\:\:m^3 \:\:\:and\:\:\:k=0.10/year$

$C_{down}=\frac{Q_{up}C}{Q_{down}+kV}=\frac{9*10^4*6}{8*10^4+0.10*2*10^5}=5.4\:\:mg/L$